An enzyme-catalysed reaction can be roughly divided into three stages:
enzyme-substrate binding, "catalysis" and product release.
"Catalysis" refers to all the steps that happen to convert substrate into
product. Sometimes, these steps are too fast to distinguish from each
other. To simplify, we sometimes refer to this whole sequence of events as
though they were just one step.

Often, but not always, that catalysis part is the rate determining step.
Product release is sort of an afterthought.

In that case, we might simplify and only consider those steps up through
catalysis.

If we do that, we find that enzyme reactions can be summarized by a relation
called the Michaelis-Menten equation, named after the early 20th century
biochemist Leonor Michaels and his collaborator, the
immensely talented artist, physician and biochemist,
Maud Menten.

The numerator in this equation should make sense -- of course the rate should
increase when we add either more substrate or more enzyme and the relationship
will depend on the speed of that product-forming step.

The denominator is a little more complicated and contains the composite constant
Km, the Michaelis-Menten constant.

This relationship can be understood most easily by examining its limits.
For instance, suppose the substrate concentration is still very low.
Perhaps it is much smaller than Km. How does that
affect the rate of the reaction?

It is useful to keep in mind that a large number added to a small number is just
a large number. One million plus one is about a million. One million
plus three is also about a million. One million plus five is pretty close
to a million. We can often ignore the small quantity in additions.
In that sense, if [S] is small, we can ignore it in the denominator, and think
of the denominator instead as "approximately Km".

We can't ignore [S] in the numerator, of course, even if it is very small.
That's because a number multiplied by a very small number also becomes a very
small number; the small number really counts when it is multiplied by something.

On the other hand, when [S] gets very large, we can ignore Km. That means
Km disappears from the denominator, leaving only [S]. At that point, the
[S] in the denominator and numerator cancels.

Remember, when [S] gets very large, the reaction has reached its maximum
possible rate, because the enzyme is saturated. Adding more substrate
doesn't speed things up, because the extra substrate just has to wait around
until an enzyme becomes available. We call that maximum rate Vmax.

The limits of the Michaelis Menten equation explains the shape of the curve
describing the rate dependence on substrate.

There is another piece of useful information we can get from this equation.
It comes at an intermediate point between the two cases we have considered so
far. What if the numerical value of Km and [S]
are exactly equal? In that case, the value of the denominator, Km
+ [S], is the same as [S] + [S].
The denominator becomes 2[S]. Just as in the limiting
case that gave us the value of Vmax, the [S]
cancels, but this time there is still a 2 in the denominator, so we get Vmax/2.

Turning that conclusion around, if we find the point on a Michaelis Menten
plot where the rate is half the maximum rate, we can drop a line down to the x
axis. The value of [S] at the intercept will be numerically the same as
the value of Km.

Vmax is the point on the y axis where the rate has
completely leveled off.

Km is the point on the x axis corresponding to y =
Vmax/2.

Problem MK5.1.

Determine Vmax and Km in each
of the following cases. Assume the units of [S] are millimoles per liter
and the units of V are moles per liter per second.

a)

b)

c)

d)

e)

Turnover Frequency and Efficiency

If we're running an experiment, we know what the total concentration of enzyme
is, because we're the ones who put it in there. That means that we can
also figure out exactly what that rate constant is for catalysis.

kcat = k2 = Vmax / [Etot]

That quantity, kcat, is sometimes referred to by biochemists
as the "turnover number". The turnover number essentially means the number
of molecules of product made by an enzyme in the specified period of time
(usually the units of kcat are expressed as s-1,
but they could also be written in min-1, etc).

In industrial catalysis, kcat is instead
referred to as the "turnover frequency", but of course it still means the same
thing. There is an important reason for this difference in terms.
The "turnover number" in industry refers to the number of molecules of product
made before the catalyst stops working. Catalyst death can occur for any
number of reasons, but you might imagine something going wrong via a side
reaction that renders the catalyst unreactive toward the substrate. This
is a very important consideration in industry. The engineer in charge of
the production plant would like to replace the catalyst with a new batch
before it stops working, to avoid an unscheduled halt in the process that
could prove very costly. They need to have an idea about when that is
likely to happen, so they need to be aware of the turnover number in this sense.

Another consideration that is sometimes useful is enzymatic efficiency.
Remember, the reaction does not depend only on the catalysis step. The
binding step also matters. The faster the catalysis step, the faster the
production of product. In addition, the greater the proportion of
substrate bound, the faster the production of product.

Combining those two ideas:

Efficiency = kcat / Km

In this relationship, Km is a stand-in for the
equilibrium constant for enzyme-substrate dissociation. It's not quite the
same thing, but it's the closest we've got. By extension, 1/Km
stands for the enzyme-substrate binding constant. The greater the binding
constant and the faster the catalysis, the more efficient the enzyme.

Note that the units of Km are concentration units (mol L-1,
for instance). The units of efficiency will therefore be something like L
mol-1 s-1.

Lineweaver-Burk Plots

The Michaelis-Menten equation is useful in other ways, too. If we take its inverse, we get
a new relationship.

That's useful because it's really an expression for a straight line. If we
plot 1/v against 1/[S], we get a straight line. The slope is Km/Vmax
and the y intercept is 1/Vmax.

Lineweaver-Burk plot gives a straight line for the rate data.

y intercept = 1/Vmax

slope = Km/Vmax

the x intercept = -1/Km, too

Problem MK5.2.

Determine the values of Vmax and Km in each of the
following cases. Assume the units of [S] are millimoles per liter and the
units of V are moles per liter per second.

a)

b)

c)

d)

e)

f)

Inhibition and Lineweaver-Burk Plots

Lineweaver-Burk plots allow additional insight into the mechanism of inhibition.
The following plot, for example, shows competitive inhibition.

Because the plot uses 1/v on the y axis, the slower reaction is actually the top
line. The bottom line is the regular reaction without an inhibitor.
Remember, the y intercept is equal to 1/Vmax. In
competitive inhibition, the inhibited reaction eventually reaches (or at least
approaches) the same Vmax as the uninhibited reaction.
The Lineweaver-Burk plot shows both lines meet the y axis at the same place.

In contrast, the following plot shows noncompetitive inhibition. Once
again, the regular line is the lower one, whereas the upper line is the
inhibited one. The two lines do not share the same y intercept, however.
However, they do share the same x intercept. That's because noncompetitive
inhibition does not directly affect binding of substrate (which is reflected in
Km), but interferes with the catalysis step (linked to Vmax).

The difference between noncompetitive and uncompetitive inhibition, very subtle
in a Michaelis-Menten plot, is quite clear in Lineweaver-Burk. The case
illustrated below is thought of as "pure" uncompetitive inhibition. Once again,
the inhibited reaction is shown by the upper line. In this case, the
inhibited and uninhibited reactions produce parallel lines in the
Lineweaver-Burk plot; that feature is actually the definition of uncompetitive
inhibition.

If you think about it, pure uncompetitive inhibition only happens under very
specific circumstances. The Km clearly differs when an
inhibitor is added; we can see that in the different x intercept. However,
the slope is the same. The slope is Km/Vmax.
That means that, because Km is different, Vmax
must differ in exactly the same way, keeping the slope the same. For
example, if Km is cut in half in the inhibited reaction,
then Vmax must also be cut in half. That would keep
the slope the same.

On the other hand, a situation in which the inhibited reaction does not give a
line parallel to the regular reaction is called "mixed inhibition". Like
uncompetitive inhibition, mixed inhibition results in changes in both Km
and Vmax. However, the changes in this case do not
scale exactly the same way as in uncompetitive inhibition.

Problem MK5.3.

Characterise each of the following graphs as representing competitive,
noncompetitive, uncompetitive, or mixed inhibition.

a)

b)

c)

d)

e)

f)

g)

The Origin of the Michaelis-Menten Equation

But where does the Michaelis-Menten relationship come from? That takes a
little bit of heavy lifting with kinetics. If you feel the need to know,
then we'll start with the approximation that the reaction essentially boils down
to two steps: substrate binding and the stuff after that. The
binding step is described as k1/k-1.
The stuff after that is summed up in k2.

The rate of product formation really depends on the rate of the elementary step
k2. That rate depends upon the amount of enzyme substrate
complex and its rate of passage through the subsequent step.

The trouble is, the enzyme substrate complex is an intermediate. We don't
know exactly how much of it we have. We might assume that, as a reactive
intermediate, the complex doesn't have much of a lifetime. It gets used up
pretty much as soon as it forms.

That assumption helps us to express the concentration of enzyme substrate
complex in terms of other things we might know more about: the enzyme and the
substrate. Of course, the enzyme and the substrate react together to make
the enzyme substrate complex. They react together with rate constant k1.

Two possible fates await the enzyme substrate complex. Either it is
released back to enzyme and free substrate, with rate constant k-1,
or else it goes on to make product, with rate constant k2.
In a steady state approximation, the enzyme substrate complex is consumed as
soon as it is formed.

Again, we don't know how much free enzyme there is. We don't know how much
enzyme-substrate complex we have. We do know how much enzyme is added at
the beginning of the kinetics experiment. We'll call that concentration [Etot],
meaning the total amount of enzyme. Some of that enzyme remains free, and
some of it is bound as enzyme-substrate complex.

It's useful to express the concentration of free enzyme as the total enzyme
minus that portion bound with substrate. That way, we'll be able to
eliminate the term for free enzyme from the rate equation. A few steps of
algebra let us express the concentration of the enzyme-substrate complex solely
in terms of the total enzyme concentration, the substrate concentration and some
rate constants.

Remember, the rate of product formation just depends on the amount of
enzyme-substrate complex and the rate constant for the catalysis step.

That collection of constants in the denominator is just a group of numbers.
It's a constant. We'll call it the Michalis-Menten constant.

That brings us back to the Michaelis Menten equation.

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's
University (with contributions from other authors as noted). It is freely
available for educational use.